如何将AST作为R中的列表获取 [英] How to get AST as a list in R

问题描述

我有描述数学公式的字符串，我想将其转换为有意义的部分列表.函数 ast _ 确实知道如何解析它，将其显示为抽象语法树，但不返回AST.我正在寻找一个确实返回树的函数.

I have strings, which describe mathematical formulas and I would like to convert it into lists of meaningful parts. The function ast_ does know how to parse it, to displays it as an Abstract Syntax Tree but does not return the AST. I am looking for a function which does return the tree.

bb <- "(media.urin_A + media.urin_B)/2"
lazyeval::ast_(rlang::parse_expr(bb))

> lazyeval::ast_(rlang::parse_expr(bb))
┗ ()
 ┗ `/
 ┗ ()
  ┗ `(
  ┗ ()
   ┗ `+
   ┗ `media.urin_A
   ┗ `media.urin_B
 ┗  2 

推荐答案

您可以使用 purrr :

getAST <- function(ee) purrr::map_if(as.list(ee), is.call, getAST)

# Example usage on expressions:
getAST( quote(log10(a+5)/b) )
# List of 3
#  $ : symbol /
#  $ :List of 2
#   ..$ : symbol log10
#   ..$ :List of 3
#   .. ..$ : symbol +
#   .. ..$ : symbol a
#   .. ..$ : num 5
#  $ : symbol b

# Example usage on strings:
getAST( str2lang("(media.urin_A + media.urin_B)/2") )
# List of 3
#  $ : symbol /
#  $ :List of 2
#   ..$ : symbol (
#   ..$ :List of 3
#   .. ..$ : symbol +
#   .. ..$ : symbol media.urin_A
#   .. ..$ : symbol media.urin_B
#  $ : num 2

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